I know how to parametrize a cone surface that has its vertex on the origin. However, how should one parametrize a cone that is sitting on the $xy$ plane, that is to say the cone has its base on $xy$ plane, radius 5, say, and its vertex is on $(0,0,h)$ say.
Can I just do $(z-h)^2=x^2+y^2$ and restrict the values of $z$? Also how does the radius come in to play?
In general, cone is defined as the place of the lines passing through a point (vertex) and resting on the points of a curve (directrix), whose points $P(x,\,y,\,z)$ can be parameterized as follows: $$\begin{cases} x = x_v + \left(x(u) - x_v\right)v \\ y = y_v + \left(y(u) - y_v\right)v \\ z = z_v + \left(z(u) - z_v\right)v \end{cases} \; \; \; \; \; \; \text{with} \; (u,\,v) \in I \times J\,,$$ where $(x_v,\,y_v,\,z_v)$ are the coordinates of the vertex, $\left(x(u),\,y(u),\,z(u)\right)$ are the parametric equations of the directrix and $I,\,J$ are subsets of $\mathbb{R}$.
In particular, if the vertex has coordinates $(0,\,0,\,h)$ and the directrix is parameterizable as $\left(5\,\cos u, \; 5\,\sin u, \; 0\right)$ with $u \in [0,\,2\pi)$, the respective cone can be parameterized as: $$\begin{cases} x = 5\,v\,\cos u \\ y = 5\,v\,\sin u \\ z = h(1-v) \end{cases} \; \; \; \; \; \; \text{with} \; (u,\,v) \in [0,\,2\pi) \times J\,,$$ where, in general, $J \equiv \mathbb{R}$, while if we want to limit the cone between the vertex and the directrix, we have $J = [0,\,1]$. Finally, if necessary, noting that: $$x^2+y^2 = 25\,v^2\,, \; \; \; \; \; \; (h-z)^2 = h^2\,v^2$$ it is also possible to derive the cartesian equation of the cone: $$x^2 + y^2 = 25\left(1-\frac{z}{h}\right)^2$$